Joint Transfer of Energy and Information in a Two-hop Relay Channel

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1 Joint Transfer of Energy and Information in a Two-hop Relay Channel Ali H. Abdollahi Bafghi, Mahtab Mirmohseni, and Mohammad Reza Aref Information Systems and Secrity Lab (ISSL Department of Electrical Engineering, Sharif University of Technology, Tehran, Iran hajiabdollahi a@ee.sharif.ed,{mirmohseni,aref}@sharif.ed arxiv: v1 [cs.it] 13 Apr 2017 Abstract We stdy the problem of joint information and energy transfer in a two-hop channel with a Radio freqency (RF energy harvesting relay. We consider a finite battery size at the relay and deterministic energy loss in transmitting energy. In other words, to be able to send an energy-contained symbol, the relay mst receive mltiple energy-contained symbols. Ths, we face a kind of channel with memory. We model the energy saved in battery as channel state with the challenge that the receiver does not know the channel state. First, we consider the problem withot any channel noise and derive an achievable rate. Next, we extend the reslts to the case with an independent and identically distribted noise in the second hop (the relay-receiver link. I. INTRODUCTION Nowadays energy consmption becomes an important design factor in commnication systems instead of traditional parameters like throghpt becase of financial reasons and environmental concerns. There wold be many eqipments with large energy consmption in next generation (5G networks, so energy efficiency will play an important rle in these networks [1]. One of alternative techniqes sed for energy consmption management is energy harvesting. Energy harvesting enables wireless networks to se environmental energies to increase energy efficiency, so reqired energy for commnication nodes in wireless networks decrease noticeably. This promising method is introdced in two main directions: (i in the first direction, energy is harvested from environmental sorces like wind and snlight. The main characteristic of this setp is the sporadic natre of the harvested energy which makes the exact analysis rather difficlt; (ii in the second direction, known as Radio freqency (RF energy harvesting, energy is harvested from the radio waves in space. This techniqe is promising with an increasing demand profile and some commercialized prodcts [2]. De to less randomness in obtaining the harvested energy, the analysis and strategies cold be simpler compared with the harvesting from the environment. The RF energy can be transferred concrrently with the information signal in a wireless system, proposed as simltaneos wireless information and power transfer (SWIPT [1], [2]. A sefl scenario in this case is to se received energy for ftre transmissions. In this scenario, the design of encoder and decoder is a new open problem, becase This work was spported in part by the Iran National Science Fondation nder Grant of the memory now appeared in the system. In addition, the existing works show an inherent trade-off in transmitting energy and information simltaneosly [2]. Ths to nderstand the interplay among energy and information and also to obtain the optimal coding strctre in this scenario, an information theoretic model works. Becase of complexity of models (especially channels with memory, fndamental limits in this system are not noticed widely. In particlar, it is not clear what are the properties of an optimal code in this system(the code which transmits information and energy simltaneosly. One of the encoding methods for the joint energy and information transfer ses a finite state Markov sorce to generate codewords [3]. The energy reception constraint can be modeled with a channel with constraint on otpt, whose capacity was derived by Gastpar in [4]. He also extended this reslt to mltiple access and Gassian relay channels [4]. The problem of transmission of optimal rate with constraint on minimm received energy was stdied by Varshney in [5], where the capacity-energy fnction was introdced and some of its featres were characterized. A channel with stochastic power restriction was stdied by Ozel and Ulks [6]. They showed that the capacity of the AWGN channel with random power available at the transmitter is the same as the capacity of an AWGN channel with an average power constraint eqal to the average recharge rate. Capacity of a mltiple access channel with constraint on received power was derived by Foladgar and Simeone [7]. They also stdied the mlti-hop channel with energy harvesting relay and introdced capacityenergy fnction. However, they did not consider a finite battery at the relay [7]. The capacity of a point to point channel with an energy harvesting transmitter was determined by Ttncogl et. al. [8], where the battery size was assmed to be one. The problem of interactive commnication with energy transfer (re-transmit the received energy with finite battery sizes (i.e., finite energy nits available in the system was stdied by Popovski et. al. [9]. They derived the inner and oter bond for a two way orthogonal channel with finite energy nits. In their system model, two nodes have a constant sm of energy nits. If a node wants to send symbol 1, it has to cost one energy nit, bt sending symbol 0 does not cost any energy nit. Similarly if a node receives symbol 1, it can save one energy nit and receiving symbol 0 does not have any energy nit [9]. One of the important challenges which

2 is not considered in [9] is that the receiver node cannot save energy of received signal entirely and an energy loss occrs. This energy loss can be modeled by assming that the receiver has to receive m energy-contained symbols for sending a symbol with energy. Another interactive scenario that can be stdied to nderstand the natre of the optimal codes in a joint information and energy transfer is relay channel. In this paper, we consider a two-hop channel with an RF energy harvesting relay, where the transmitter jointly transfers information and energy to the relay. The harvested energy at the relay is sed to re-transmit the data to the receiver. We assme finite battery size at the relay. The energy loss in transmitting energy is modeled with a fixed deterministic redction in energy. In fact, the relay mst receive mltiple energy-contained symbols to be able to send one energycontained symbol. These limitations at the relay trn the problem to the transmission over a channel with states, where the state shows the energy level at the relay s battery. Hence, we face a kind of channel with memory. Ths, the main qestions are which rates wold be achievable in these models and what the strctre of the coding techniqes are that achieve those rates. One of the main challenge in the code design is to make the receiver be able to decode the message withot knowing the seqence of states. We model the energy stored in relay s battery as channel state with the challenge that the receiver does not know the channel state. First, we consider the problem withot any channel noise and derive an achievable rate. We propose a new block Markov coding based achievability scheme in which the random codebooks are generated for each state. We show that the received codewords in the receiver form Markov sorces. We se this property and Asymptotic Eqipartition Property (A.E.P Theorem [10] to propose a new decoding strategy which does not need state seqence at the receiver. Next, we consider the problem with an independent and identically distribted noise in the second hop (the relay-receiver link and find a coding scheme which works in the noisy condition. Ths, we extend or achievable reslts to the noisy case, where we modify the decoding strategy at the receiver by sing typical decoding with the capability of decoding withot knowing state seqence. II. SYSTEM MODEL We consider a binary and noiseless two-hop relay channel illstrated in Fig. 1, in which the relay node has energy restriction (i.e., the relay mst harvest energy from its received signal to be able to transmit. Also, we assme a finite battery at the relay which can save finite nmber of energy nits. Ths, the transmitted symbol depends on the harvested energy from received symbols. Notation: Upper-case letters (e.g., X denote Random Variables (RVs and lower-case letters (e.g., x their realizations. The probability mass fnction (p.m.f of a RV X with alphabet set X is denoted by p X (x; occasionally, sbscript X is omitted. X j i indicates a seqence of RVs (X i, X i+1,..., X j ; we se X j instead of X j 1 for brevity. The channel inpts at the transmitter and the relay are shown by {X 1,1, X 1,2, X 1,3,...} and {X 2,1, X 2,2, X 2,3,...}, respectively. The otpt at the relay and the receiver s are denoted by {Y 2,1, Y 2,2, Y 2,3,...} and {Y 3,1, Y 3,2, Y 3,3,...}, respectively. The channel inpt at the transmitter and the channel otpts at the relay and the receiver have binary alphabets, i.e., X 1 = Y 2 = Y 3 = {0, 1}. The alphabet of channel inpt at the relay is shown by X 2 (will be introdced later. {U 1, U 2, U 3,...} show the level of battery storage in the i-th transmission which are sed to model the channel state. S i are considered as: S 1 = (U 1, U 2, S 2 = (U 2, U 3, S 3 = (U 3, U 4,... which are sed in frther proofs. π denotes the steady state probability of the -th state of the Markov chain. In or system model transmitting symbol 1 needs m energy nits, while symbol 0 can be sent with no energy (the transmitting symbol 1 costs m energy nits for the relay. However, receiving symbol 1 at the relay charges the battery with only one energy nit. This consideration shows the energy loss in the channel. The transmitter does not have energy restriction and it can transmit any symbol in each channel se. The transmitter sends a message M [1, 2 nr ] to the relay node in n channel ses (by transmitting X n 1. Then, the relay decodes this message and sends it to the receiver (by transmitting X n 2. The X 2, is the set of symbols which can be transmitted by relay when channel state is. The channel state shows the energy nits stored in the relay s battery and U is the maximm battery size. Energy restriction in the relay node is described as: < m X 2, {0} (1 m X 2, {0, 1} (2 where [0 : U]. We call this system as noiseless twohop relay channel with finite battery (Noiseless THRC-FB. Becase of the noiseless channel property, the otpts are eqal to inpts in each hop, i.e., Y 2 = X 1, Y 3 = X 2, where Y 2, Y 3 are the received symbols at the relay and the receiver. The main difficlty here is that the system has memory de to the energy restriction and finite battery size at the relay. Or approach is to model the energy nits (in the relay s battery as the state of the system. Also, we assme that the receiver does not know the state seqence, which is another difficlty we face. The state diagram of the channel is shown in Fig. 2. As seen in Fig. 2, when the battery is in state in the crrent transition, the following cases occr in the state diagram: iif the relay receives symbol 1 and transmits symbol 0, the state in the next transition wold be +1 (except when = U, where the next state won t change; iiif the relay receives symbol 0 and transmits symbol 1, the state in the next transition wold be m (in this case we mst have > m 1, otherwise it does not occr; iiiif the relay receives symbol 1 and transmits symbol 1, the state in the next transition wold be m + 1 (in this case we mst have > m 2, otherwise it does not occr; ivif the relay receives symbol 0 and transmits symbol 0, the next state wold be the same as the crrent state. Now, to insert the noise in to the problem, we consider a binary memoryless channel between the relay and the

3 Fig. 1. Noiseless two-hop relay channel with finite battery (Noiseless THRC- FB Fig. 2. State diagram of noiseless THRC-FB receiver (the second hop. In this case, noise has the ability to change the transmitted symbol, and of corse its energy, randomly. This means that if channel converts symbol 1 into symbol 0, it wold not have energy and if channel converts symbol 0 into symbol 1, it wold contain one energy nit. However, this change does not affect the channel state diagram, and ths it is not important in the channel memory. This system model is illstrated in Fig.3. All of other considerations are exactly as same as previos model. We call this system as noisy two-hop relay channel with finite battery (Noisy THRC-FB. Encoding and decoding fnctions depend on battery size U, so we have to inclde this parameter in or code definition. A (2 nr, n, U code for (Noiseless/Noisy THRC-FB consists of a message set [1 : 2 nr ], an encoder fnction which maps m [1 : 2 nr ] to x n 1 (m, a set of relay encoder fnctions which map each past received seqence x i 1 1 to x 2,i (x i 1 1 for i [1 : n], and a decoder fnction which estimates ˆm from the received seqence y3 n in receiver. We define the average probability of error P e (n = P {M ˆM}. A rate R is achievable for (Noiseless/Noisy THRC-FB, if there exists a (2 nr, n, U code sch that lim P e (n = 0 n III. NOISELESS THRC-FB In this section we propose an achievable rate for the noiseless THRC-FB. First we provide a lemma (to be sed in the achievability proof, where we state a sfficient condition for Fig. 3. Noisy two-hop relay channel with finite battery (Noisy THRC-FB- noise between the relay and the receiver existence of steady state probabilities in a finite state Markov chain. Next, we prove an achievability theorem based on block Markov coding. We sed sperposition coding for codebook generation for each state. The novel part of or work is in the decoding at the receiver withot knowing the state seqence. We se backward decoding and A.E.P Theorem [10, Theorem 6.6.1] for message decoding in receiver. Lemma 1: Consider an indecomposable Markov chain with r possible states. The steady state probabilities exist, if there exists a state S in the state diagram which is accessible from itself in one transition (the probability of retrning to itself in next transition is nonzero. Proof: It is known that a sfficient condition for existence of the steady state probabilities is that there exist a state Ŝ and a positive nmber n sch that beginning from any state we can reach state Ŝ in n steps [10, Theorem ]. Now, we show these conditions hold by choosing Ŝ to be S and n to be maximm distance between S and any other states in state diagram. Let S k be the states which has distance k from S and m be the maximm distance, i.e., m = max k. One can reach S from S k in l steps, where l is an arbitrary integer nmber that l k. Becase, after first arrival to S, we can stay there (de to the assmption of lemma. Therefore, beginning from any state we can reach S in at least m steps. Theorem 1: The following rate, R, is achievable for Noiseless THRC-FB: R < max min{ U π H(X 2, π H(X X2 1 } p(x 1,x 2 (3 where p(x 1, x 2 is chosen sch that the state diagram of channel satisfies assmptions of Lemma 1. This means that the state diagram is indecomposable and there exists a state S in the state diagram which is accessible from itself in one transition. For simplicity, from now on, we call this class of p.m.f.s indecomposable. { Note that for [0 : m 1] we mst 1 x2 = 0 have p(x 2 =. 0 x 2 = 1 Proof: Or scheme ses block Markov coding, where the B blocks of transmissions (each of n symbols are sent to the relay node to transmit a seqence of B 1 independent and identically distribted (i.i.d messages M b, b [1 : B 1] while the message of the last block is deterministic. Similarly, the relay node sends B blocks to the receiver in which the message of the first block is deterministic and the messages of the remaining blocks are the same as the transmitter s message with one block delay. At the end of each block, the relay decodes the message and sends it to the receiver in the next block. Ths, the relay sends a deterministic message in the first block and the transmitter sends a deterministic message in last block. Since the state space is finite, we can control the initial state in each block with at most U transmissions. Ths, we assme that the initial state in each block can be adjsted and for simplicity we do not contain these U transmissions in or frther discssions. In fact, by inclding these transmissions,

4 Fig. 4. Sperposition coding for each state with joint p.m.f p(x 1, x 2 each block contains n + U bits instead of n bits. Codebook generation: For each state U, fix an indecomposable p.m.f p(x 1, x 2. Now, for each state U and for each block b [1 : B 1], we generate randomly and independently 2 nr seqences x n+δ 2 (m b 1, where m b 1 [1 : 2 nr ], each according to n+δ p X2 (x 2,i. For each m b 1 [1 : 2 nr ], we generate randomly and conditionally independently K seqences x n+δ 1 (m,b, m b 1, where m,b [1 : K ], each according to n+δ p X1 X 2 (x 1,i x 2,i (m b 1, where K is the size of the transmitter s codebook of state and m,b is the message of the transmitter s codebook of state in block b. We have m 0 = m B = 1. In fact, we do a kind of rate splitting in transmitter in which K = 2 nr and R is the rate of each codebook. The codewords are shown in Fig. 4. In addition, we generate 2 nr i.i.d random variables U(m b 1, where m b 1 [1 : 2 nr ], with p.m.f π. These random variables will be sed as initial state of channel in block b. We remark that only the first n bits (in each codeword contain message and we find n for each state sch that the error probability tends to zero. Other δ bits are generated to protect channel s statistical properties (Markovity from change. This means that we generate δ joint random bits from the p.m.f p(x 1, x 2 for each message set (m,b, m b 1. Ths, if n bits of codeword of state are sent completely before the codewords of other states, sending these δ bits wold prevent the changing of statistical properties of channel state diagram. δ can be chosen as large as n min(n to satisfy the above condition. The transmission strategy is described in the following. Encoding (at the beginning of block b U Transmitter: To send message m b [1, K ] in block b, transmitter maps the message into a message vector [m 0,b, m 1,b,..., m U,b ], m,b [1 : K ]. Then, it selects the codeword x n+δ 1 (m,b, m b 1 from the codebook corresponding to state. In addition, we set a vector as: [l 0,b = 1, l 1,b = 1,..., l U,b = 1]. For encoding in block b, Fig. 5. Flowchart of encoding block b in transmitter and relay starting from the beginning of the block with 1 as initial state, we send x (m 1 1,l 1,b 1,b, m b 1 and we set l 1,b = l 1,b +1. We repeat this procedre for next transmissions. Encoding procedre is shown completely in Fig 5. This procedre is similar to the encoding introdced in [9]. Relay: The relay sends message m b 1 in block b, so it selects codeword m b 1 from each codebook as the message of that codebook and chooses U(m b 1 as initial state. The next stages are the same as the encoding at the transmitter and are shown in Fig. 5. Decoding (at the end of block b Relay: The state seqence in block b, n b, is known at the relay, becase it knows its battery storage. Ths, for each [0 : U], the relay makes a set A = {i i = }. If A n, the relay looks for an ˆm,b = {m x2,k,b (m,b = x 2,b,ik }, k [1 : n ] and i 1 i 2... i n. This procedre is shown in Fig. 6. In the error probability analysis, we find the conditions which garantee the ˆm,b to be niqe. Then, the relay forms the vector [ ˆm 0,b, ˆm 1,b,..., ˆm U,b ] and by the inverse of mapping sed in encoding, it can decode ˆm b. This procedre is like the decoding introdced in [9]. Receiver: The receiver ses backward decoding. In the last block, the transmitted message is fixed, i.e., m B = 1. Assme that the relay transmits the message m B 1 in last block, where m B 1 [1 : 2nR ]. The receiver rns the flowchart shown in Fig. 5 for each m B 1 and comptes the seqence generated at the relay for each m B 1. We call these seqences Xn 2 (m B = 1, m B 1.Indeed, Xn 2 (m B = 1, m B 1 is the seqence which is received at the receiver if the relay sends message m B 1 in last block. To perform the above decoding, the receiver does not need the state seqence, becase: 1the codebooks and initial states are shared; 2there is not any noise; 3as we see in the flowchart of Fig. 5, since the transmitter s message (in last block m B = 1, relay s message (m B 1 and the initial

5 Fig. 6. Flowchart of decoding block b in relay state are known, after every transmission the next state can be determined and X2 n (m B = 1, m B 1 cold be derived for each m B 1 [1 : 2nR ]. Then, the receiver looks for a niqe ˆm B 1, sch that its compted seqence is eqal to the received seqence. We show that 2 nr probable received seqences are not eqal with probability 1. When message m B 1 is decoded, the transmitter s message in block B 1 is known and the above procedre can be repeated to decode the previos blocks messages. Error probability analysis: The probability of error is pper bonded by the sm of probabilities of error in the relay and the receiver. The error events at the relay in each block are: ε (1 = Relay does not receive the codeword of at least one of the codebooks (corresponds to a state completely. We define ε (1 as the event in which the relay does not receive the codeword of codebook completely. ε (2 = There are more than one eqal codewords with received seqence in at least one of codebooks. We define ε (2 as the event in which there are more than one eqal codewords with the received seqence of codebook. where it is seen that P (ε (i P (ε (i, i {1, 2}. First, we consider the ε (1. Recall that the state seqence has steady state by Lemma 1. In addition, based on a reslt in [10, Theorem ], in a finite Markov chain with steady state, the relative freqency of being in a state converges to the steady state probability π, in probability. Ths, if we choose n = n(π ɛ, the event ε (1 does not occr with probability 1 and P (ε (1 goes to zero for large enogh n. Based on Lemma (2, the probability of the second error event (ε (2 goes to zero if: log(k n < H(X 1 X2 ɛ (4 Lemma 2: Fix a joint p.m.f p(, x and generate a random seqence U n according to n p U ( i, then generate randomly and conditionally independently 2 nr seqences X n (m, m [1 : 2 nr ], each according to n p X U (x i i. If we have R < H(X U, then probability of the event {X n (m = X n (m, m m }, wold tend to zero. Proof: Based on packing lemma [11], if the seqence X n (1 is passed throgh a discrete memoryless channel n p Y X (y i x i and the seqence Y n is constrcted and if R < I(X; Y U, then we wold have P ((U n, X n (m, Y n A n ε (p(, x, y 0, m 1. Now we consider a p(y x sch that p(y = x x = 1, p(y x x = 0, so we obtain Y n = X n (1. Now we determine A n ε (p(, x, y. By definition we have: A n ε (p(, x, y = {(x n, n, y n π(x,, y x n, n, y n p(, x, y εp(, x, y, U, x X, y Y} (5 in which π(x,, y x n, n, y n is the percentage of repetition of, x, y in the seqence x n, n, y n. For conditional distribtion described above, we have p (, x, y x = 0, p (, x, y = x = p (, x p (y x = p (, x, and ths: π (, x, y x n, x n, y n = 0 (6 π (, x, y = x n, x n, y n = π (, x n, x n (7 By inserting (6 and (7 in (5, we get: A n ε (p (, x, y = {( n, x n, y n ( n, x n A n ε (p (, x, x n = y n } (8 which reslts in: P ((U n, X n (m, Y n A n ε (p(, x, y = P ((U n, X n (m A n ε (p(, x X n (m = X n (1 So by packing lemma [11] for m 1, we have: P ((U n, X n (m, Y n A n ε (p(, x, y < ε P ((U n, X n (m A n ε (p(, x X n (m = X n (1 < ε And by complementing above eqation we have: P ((U n, X n (m / A n ε (p(, x X n (m X n (1 > 1 ε (9 we apply nion bond to obtain: P ((U n, X n (m / A n ε (p(, x X n (m X n (1 P ((U n, X n (m / A n ε (p(, x + P (X n (m X n (1 (10 In addition, by Joint A.E.P Theorem [12, Theorem 7.6.1], we have: P ((U n, X n (m / A n ε (p(, x < ε (11 By combining (9-(11 we obtain: P (X n (m X n (1 > 1 ε ε = 1 ε

6 So by packing lemma [11] and discssions given in the first paragraph of proof and the fact X = Y, if we have: R < I (X; Y U = I (X; X U = H (X U then we can derive ineqality (12: P (X n (m = X n (1 < ε, m 1 (12 Where ε cold take every small vale for large enogh n. Repeating this argment for each X n (m, m [1 : 2 nr ], completes the proof. If we sbstitte n = n(π ɛ in (4, we obtain: n = n (π ɛ n < n (π 1 n > 1 n (π R = log U 2 K = n R < < π log 2 K nπ (13 π log 2 K n (14 π H(X 1 X 2 (15 where eqation (14 is derived by eqation (13 and eqation (15 is derived by eqations (4 and (14. For error probability analysis in receiver, recall that in the last block, the transmitter s message is deterministic and so the receiver derives X2 n (m B = 1, m B 1, m B 1 [1 : 2 nr ]. Since we assme that relay sends X2 n (m B = 1, m B 1 = 1, If at least one X2 n (m B = 1, m B 1 1 becomes eqal to X2 n (m B = 1, m B 1 = 1, the error occrs in the receiver. Lemma 3: For each m B 1 [1 : 2nR ], the X2 n (m B = 1, m B 1 is an independent reglar Markov sorce. Proof: Independency can be easily dedced from the codebook generation becase the initial states and all codewords in each codebook are generated independently. To show that these seqences are reglar markov sorces, we define a new Markov chain as: denoting the state seqence as U 1, U 2, U 3,..., the new Markov chain is S 1 = (U 1, U 2, S 2 = (U 2, U 3, S 3 = (U 3, U 4,..., as we described in section (II. Since X 2i is determined by U i, U i+1, we have X 2,i = f(s i, where f is a deterministic fnction. Ths, X 2,i is a markov sorce. Moreover, the assmptions of Lemma 1 are also satisfied by the new Markov chain S i and so steady state probabilities exist. This shows that X 2,i is a reglar Markov sorce. Since a reglar Markov sorce is ergodic [10, Theorem 6.6.2], X 2,i satisfies conditions of Asymptotic EqiPartition Property (A.E.P Theorem [10, Theorem 6.6.1] which states that if we have 2 nr i.i.d Markov sorces with entropy rate H{X}, these Markov sorces are not eqal with probability 1 when R < H{X}. Now, we derive the entropy rate of X 2,i. Since X 2,i is not nifilar, we cannot se entropy rate of nifilar Markov sorces (a nifilar Markov sorce is a Markov sorce in which the present state U n and the present otpt X n compte the next state U n+1. We se the followings (for simplicity, the index B for block nmber is omitted from the eqations: H(X 2,n X 2,n 1,..., X 2,1 H(X 2,n U n, X 2,n 1,..., X 2,1 (16 = H(X 2,n U n = H(X 2,1 U 1 = π H(X 2 (17 The ineqality (16 follows from the fact that conditioning does not increase the entropy and the eqation (17 holds since conditioning on U n, the distribtion of X 2,n is determined independent of X 2,n 1,..., X 2,1 and or processes are stationary. Therefore, if R < U π H(X 2, then the generated seqences are not eqal with probability 1 by A.E.P Theorem. This completes the proof. Note that to find K, first we have to solve optimization problem in eqation (3, so we can determine p(x 1, x 2. Next, we calclate H ( X 1 X 2 and we consider K < H ( X 1 X2. IV. NOISY THRC-FB Theorem 2: The following rate, R, is achievable for Noisy THRC-FB: R < max min{ U π I(X 2 ; Y 3, π H(X X2 1 } p(x 1,x 2 (18 where P (x 1, x 2 is an indecomposable { p.m.f and for 1 x2 = 0 [0 : m 1] we mst have p(x 2 =. 0 x 2 = 1 Proof: All proof steps of this theorem are exactly the same as the proof of Theorem 1, except the decoding at the receiver and the error probability analysis in receiver. Ths, we only highlight the differences. Receiver decoding: Receiver ses backward decoding. As we showed for noiseless THRC-FB, the relay comptes X2 n (m B = 1, m B 1 for each m B 1 [1 : 2nR ]. Then, the receiver looks for ˆm B 1 which satisfies (X2 n (m B = 1, ˆm B 1, Y3,B n τ ε (n, where the Y3,B n is received seqence in receiver in block B. Error probability analysis in receiver: As we showed in Lemma 2, X2 n (m B = 1, m B 1 are independent and identical Markov sorces (for m B 1 [1 : 2 nr ]. If the relay sends the seqence X2 n (m B = 1, m B 1 = 1 in block B, the received seqence in the receiver becomes independent of the other seqences X2 n (m B = 1, m B 1 1. Ths, we calclate probability of the event in which (X2 n (m B = 1, ˆm B 1 1, Y3,B n τ ε (n (for simplicity, the index B for block nmber is omitted from eqations: P e = P (X2 n = αp (Y3 n = β (α,β A n ε 2 n(h{x2}+ε 2 n(h{y3}+ε 2 n(h{x2,y3} ε (19 So the pper bond on the error probability is: P error 2 nr 2 n(h{x2}+h{y3} H{X2,Y3} 3ε (20

7 Ths the error probability tends to zero, if: 1 n R < H { X 2 } + H { Y3 } H { X2, Y 3 } = lim n (I (Xn 2 ; Y3 n (21 In addition, Y3 n is stationary, so we have: 1 lim n n H (Y 3 n = lim H(Y 3,n Y 3,n 1, Y 3,n 2,..., Y 3,1 n The term in the left hand side can be written as: H(Y 3,n Y 3,n 1,..., Y 3,1 H(Y 3,n U n, Y 3,n 1,..., Y 3,1 (22 = H(Y 3,n U n = H(Y 3,1 U 1 = π H(Y 3 (23 The reasons for (22 and (23 are exactly the same as the ones for (16 and (17. On the other hand, we have a binary memoryless channel which satisfies: H (Y n 3 X n 2 = which can be contined as: H (Y 3,k X 2,k = x = x = = n H (Y 3,k X 2,k k=1 H (Y 3,k X 2,k = x p (X 2,k = x H (Y 3,k X 2,k = x π p (X 2,k = x (24 π H (Y 3 X 2 = x p (X 2 = x x π H ( Y 3 X2 (25 (26 where (24 is de to the law of total probability and the eqation (25 holds thanks to the stationarity of X 2,k, Y 3,k. By combining (21, (23 and (26, we derive: 1 n (I (Xn 2 ; Y n 3 π I ( Y 3 ; X 2 (27 Now, based on (20 and (27, we see that if R < π I ( Y 3 ; X 2, then the error probability in receiver tends to zero. Hence proof is complete. V. DISCUSSIONS AND CONCLUSIONS We stdied a two-hop channel with an RF energy harvesting relay (with finite battery size, where the transmitter jointly transfers information and energy to the relay. Modeling the energy level at the relay s battery with states, we propose the achievability schemes for the channel with memory, where the main challenge was the nknown state at the receiver. Or proposed schemes work for the noiseless channel and the channel with noisy second hop. Noisy first hop: By considering the noise in the first hop (between the transmitter and relay, the transmitter does not know the relay s battery level. Ths, the state is not available to the transmitter and as a reslt the receiver cannot compte the possible transmitted seqences of relay (for each message. Therefore, the proposed schemes are not readily extended to this case. Designing appropriate coding schemes for this channel is or ongoing research work. Upper bond: De to the channel memory, the problem of finding a tight oter bond for this system model cannot be tackled by sing standard ineqalities sed in converse proofs. REFERENCES [1] S. Bzzi, C.-L. I, T. E. Klein, C. Yang, H. V. Poor, and A. Zappone, A Srvey of Energy-Efficient Techniqes for 5G Networks and Challenges Ahead,, IEEE J. Sel. Areas Commn., vol. 34, no. 4, pp , Apr [2] X. L, P. Wang, D. Niyato, D. I. Kim, and Z. Han, Wireless networks with RF energy harvesting: a contemporary srvey,, IEEE Commnications Srveys and Ttorials, vol. 17, no. 2, pp , [3] A. M. Foladgar, O. Simeone, and E. Erkip, Constrained Codes for Joint Energy and Information Transfer,, IEEE Trans. on Commn., vol. 62, no. 6, pp , Jne [4] M. Gastpar, On capacity nder receive and spatial spectrm-sharing constraints,, IEEE Trans. Inf. Theory, vol. 53, no. 2, pp , Feb [5] L. R. Varshney, Transporting information and energy simltaneosly,, in Proc. IEEE Int. Symposim on Inform. Theory, pp , [6] O. Ozel and S. Ulks, Information-theoretic analysis of an energy harvesting commnication system,, in Proc. IEEE Int. Symposim on PIMRC, pp , [7] A. M. Foladgar, and O. Simeone, On the Transfer of Information and Energy in Mlti-User Systems,, IEEE Wireless Commn.Lett., vol. 16, no. 11, pp , Sept [8] K. Ttncogl, O. Ozel, A. Yener, and S. Ulks, Binary energy harvesting channel with finite energy storage,, in Proc. IEEE Int. Symposim on PIMRC, pp , [9] P. Popovski, A. Foladgar, and O. Simeone, Interactive joint transfer of energy and information,, IEEE Trans. on Commn.,vol. 61, no. 5, pp , May [10] R. B. Ash, Information Theory. Interscience, New York, [11] A. El Gamal and Y. H. Kim, Network information theory. Cambridge University Press, [12] T. Cover and J. A. Thomas, Elements of Information Theory. Wiley- Interscience, 2006.